PAT甲级训练刷题代码记录

刷题链接：https://www.patest.cn/contests/pat-a-practise

1001

 #include <iostream>
 #include <stdio.h>

 using namespace std;
 int temp[];
 int main()
 {
     int a,b,sum,index;
     while(scanf("%d%d",&a,&b)!=EOF){
         sum=a+b;
         if(sum<){
             printf("-");
             sum=-sum;
         }
         index=;

         while(sum>=){
             temp[index++]=sum%;
             sum=sum/;
         }
         printf("%d",sum);
         for(int i = index-;i >= ;i--){
             if(temp[i]>)
             printf(",%d",temp[i]);
         else if(temp[i]>)
             printf(",0%d",temp[i]);
         else
             printf(",00%d",temp[i]);
         }
         printf("\n");
     }
     return ;
 }

1001

数据结构初级入门第一题：https://pta.patest.cn/pta/test/17/exam/4/question/260

 #include <iostream>
 #include <stdio.h>

 using namespace std;

 int main()
 {
     int a,h,ans,i,j,temp;char c;
     while(scanf("%d %c",&a,&c)!=EOF){
         for(h=;a+>=*h*h;h++);
         if(a==)h=,ans=;
         else ans=a-*(h-)*(h-)+;
         h=(h-)*+;
         //printf("h is %d and ans is %d\n",h,ans);
         //下面开始输出
         temp=h;
         for(i=;i<h;i++){
             if(i<(h+)/){
                 for(j=;j<i;j++)printf(" ");
                 for(j=;j<h-i*;j++)printf("%c",c);
                 printf("\n");
             }else{
                 for(j=;j<h-i-;j++)printf(" ");
                 for(j=;j<*i+-h;j++)printf("%c",c);
                 printf("\n");
             }
         }
         printf("%d\n",ans);
     }
     return ;
 }

打印沙漏

第二题：思路是素数打表法，请注意做表的时候请将空间开到足够大，否则会出现越界错误。同时相对ACM来说PAT的OJ是有bug的，这也就是说明不管题目你是否会写，请得分，很简单的，至少都有分数的，加油。

#include <iostream>
#include <stdio.h>
using namespace std;

#define NMAX 100000
int num[NMAX];
int main()
{
    int i,j;
    //思路：采用素数打表法对素数进行处理
    for(i=;i < NMAX-;i++){
        num[i]=;
    }
    for(j=;j < NMAX-;j++){
        if(num[j]==)for(int k =*j;k<NMAX-;k=k+j)num[k]=;
    }
    //上边首先是对表进行初始化

    int n,count;
    while(scanf("%d",&n)!=EOF){
        count=;
        for(i=;i<n-;i++){
            if(num[i]==&&num[i+]==)count++;
        }
        printf("%d\n",count);
    }

    return ;
}

第三题：佷常见的一个题目，但是要考虑到特殊的数据集的时候，你会发现你以前不严谨的代码错误百出。还好是PAT对于结果会给你一定的提升，请考虑提高数据集的复杂性，将问题考虑全面

#include <iostream>
#include <stdio.h>

using namespace std;
int num[];

void reserve(int *,int,int);

int main()
{
    int i,n,offset;
    while(scanf("%d%d",&n,&offset)!=EOF){
        if(n==)continue;
        offset%=n;
        for(i=;i<n;i++)scanf("%d",&num[i]);
        reserve(num,,n-offset-);
        reserve(num,n-offset,n-);
        reserve(num,,n-);

        for(i=;i<n;i++)
            if(i!=n-)
        printf("%d ",num[i]);
        else
            printf("%d",num[i]);
        printf("\n");
    }
    return ;
}

void reserve(int *a,int b,int c){
    int temp;
    if(b>=c)return;
    for(int i=b;i <= b+(c-b)/;i++){
        temp=a[i];
        a[i]=a[c-(i-b)];
        a[c-(i-b)]=temp;
    }
}

第四题：这个题真的是靠着PAT的提示给整出来的，这个编译器的是好到爆，希望考试要有这样一个效果就好了

这个题目要特别注意！！！进位问题

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
using namespace std;
#define NMAX 100
char num1[NMAX],num2[NMAX];
int demo1[],demo2[];

int main()
{
    int i,j,add,count;bool flag;
    while(scanf("%s",num1)!=EOF){
        flag=true;
        for(i=;i<;i++)demo1[i]=demo2[i]=;
        for(i=;i<strlen(num1);i++)demo1[num1[i]-'']++;

        add=count=;
        for(i=strlen(num1)-;i>=;i--){
            j=num1[i]-'';
            j*=;
            num2[count++]=(j+add)%+'';
            add=(j+add)/;
        }
        if(add!=)num2[count++]=add%+'';
        num2[count]='\0';
        for(i=;i<strlen(num2);i++)demo2[num2[i]-'']++;

        for(i=;i<;i++){
            if(demo1[i]!=demo2[i]){
                flag=false;
                break;
            }
        }
        if(flag)printf("Yes\n");
        else printf("No\n");

        for(i=count-;i>=;i--)printf("%c",num2[i]);
        printf("\n");

    }
    return ;
}

第五题：终于是通关了，这个题目用的最笨的一种方法搞定的。同时也是因为自己没有理解题意，数据量太大，没法调试，卡壳了好久。参考了网上的代码后修改自己的把题目弄出来了。也是，代码这种东西满分不易，且行且珍惜。

#include <iostream>
#include <stdio.h>

using namespace std;
#define MAX 60
#define COUNT 54
int place[MAX];
int temp_num[MAX];
int place_num[MAX];

int main()
{
    int n,temp;
    char* ans[]={
        "TEMP","S1","S2","S3","S4","S5","S6","S7","S8","S9","S10","S11","S12","S13",
        "H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13",
        "C1","C2","C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13",
        "D1","D2","D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13",
        "J1","J2"};
    while(scanf("%d",&n)!=EOF){
        for(int i =;i <= COUNT;i++){
            scanf("%d",&temp_num[i]);
            place[i]=place_num[i]=i;
        }

        for(int j=;j <= n;j++){

            for(int t=;t<=COUNT;t++){
                place_num[temp_num[t]]=place[t];
            }
            for(int t=;t<=COUNT;t++){
                place[t]=place_num[t];
            }
        }

        //最后是输出结果

        for(int i = ;i<=COUNT;i++)
            if(i!=COUNT)printf("%s ",ans[place_num[i]]);
            else printf("%s",ans[place_num[i]]);
        printf("\n");
    }

    return ;
}

参考代码：

 参考代码
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <cmath>
 #include <iostream>
 using namespace std;
 char f[]={'S','H','C','D','J'};
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int time;
     scanf("%d",&time);
     int card1[],card2[],i,j;
     for(i=;i<=;i++){
         card1[i]=i;
     }
     int order[];
     for(i=;i<=;i++){
         scanf("%d",&order[i]);
     }
     for(i=;i<=time;i++){
         for(j=;j<=;j++){
             card2[order[j]]=card1[j];
         }
         for(j=;j<=;j++){
             card1[j]=card2[j];
         }
     }
     printf("%c%d",f[(card1[]-)/],card1[]%==?:card1[]%);
     for(i=;i<=;i++){
         printf(" %c%d",f[(card1[i]-)/],card1[i]%==?:card1[i]%);
     }
     printf("\n");
     return ;
 }

牛客网二叉搜索树：https://www.nowcoder.com/practice/3d6dd9a58d5246f29f71683346bb8f1b?tpId=40&tqId=21455&tPage=7&rp=7&ru=/ta/kaoyan&qru=/ta/kaoyan/question-ranking

参考博客：http://www.cnblogs.com/rickhsg/p/3650875.html

自己的题解：

 #include <stdio.h>
 #include <string>
 #include <string.h>
 #include <stdlib.h>

 typedef struct Node{
     int key;
     struct Node *left;
     struct Node *right;
 }Node;
 ///申请一个树的根节点，并将其初始设置为空
 Node * Tree=NULL,* p,* temp_p;
 char tempa[],tempb[],temp_str[];

 int length,i;
 ///先将对应的二叉树构造出来
 void Creat(char* temp){
         for(i=;i<strlen(temp);i++){
                 p=(Node *)malloc(sizeof(Node));
                 p->key=temp[i]-'';
                 p->left=p->right=NULL;

                 if(Tree==NULL){
                     Tree=p;
                 }else{
                     temp_p=Tree;
                     while(){
                         if(temp_p->key==p->key)break;
                         else if(temp_p->key<p->key){
                                 if(temp_p->right==NULL){
                                     temp_p->right=p;
                                     break;
                                 }else
                                     temp_p=temp_p->right;
                             }else
                                 if(temp_p->key>p->key){
                                     if(temp_p->left==NULL){
                                     temp_p->left=p;
                                     break;
                                 }else
                                     temp_p=temp_p->left;
                             }
                         }
                     }
                 }
 }

 void pre_order(Node* Tree){
     ///使用前应当将长度设置为0初始化
     if(Tree==NULL)return;
     else{
         temp_str[length++]=(char)(Tree->key+'');
         //printf("+%c",temp_str[length-1]);
         pre_order(Tree->left);
         pre_order(Tree->right);
     }
 }
 int main()
 {
     int T;
     while(scanf("%d",&T)!=EOF){
         if(==T)break;
         scanf("%s",tempa);
         Tree=NULL;
         ///先将对应的二叉树构造出来
         Creat(tempa);

         ///对构造好的排序二叉树进行前序遍历进行唯一性标识
         length=;
         pre_order(Tree);
         strcpy(tempa,temp_str);

         ///上边已经将二叉树构造好了，现在只要进行一个判断的工作就ok
         while(T--){
             scanf("%s",tempb);
             Tree=NULL;

             ///先将对应的二叉树构造出来
             Creat(tempb);

         ///对构造好的排序二叉树进行前序遍历进行唯一性标识
         length=;
         pre_order(Tree);
         strcpy(tempb,temp_str);

             if(strcmp(tempa,tempb))printf("NO\n");
             else printf("YES\n");
     }
     }
     return ;
 }

题解思路：https://www.nowcoder.com/questionTerminal/3d6dd9a58d5246f29f71683346bb8f1b